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Let f:(-pi/2,pi/2)-> RR be given by f(...

Let `f:(-pi/2,pi/2)-> RR` be given by `f(x) = (log(sec x + tan x))^3` Then which of the following is wrong?

A

`f(x)` is an odd function

B

`f(x)` is a one-one function

C

`f(x)` is an onto function

D

`f(x)` is an event function

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`f(x)=(log(secx+tanx))^(3) AA in (-(pi)/(2),(pi)/(2))`
` :. f(-x)=(log(secx-tanx))^(3) `
`=("log"(1)/(secx+tanx))^(3)`
`=(-log(secx+tanx))^(3)`
`= -(log(secx+tanx))^(3)`
`= -f(x)`
Hence, `f(x)` is odd function.
Let `g(x)=secx+tanx AA x in (-(pi)/(2),(pi)/(2))`
So, `g'(x)=secx(secx+tanx) gt 0 AA x in (-(pi)/(2),(pi)/(2))`
`implies g(x)` is one-one function
Hence, `(log_(e)(g(x)))^(3)` is also one-one function.
And `g(x) in (0,oo) AA x in (-(pi)/(2),(pi)/(2))`
`g(x)=secx+tanx`
`=(1+sinx)/(cosx)=(1-cos((pi)/(2)+x))/(sin((pi)/(2)+x))=tan((pi)/(4)+(x)/(2))`
Now
`x in (-(pi)/(2),(pi)/(2))`
`implies (pi)/(4)+(x)/(2) in (0,(pi)/(2))`
`implies tan((pi)/(4)+(x)/(2)) in (0,oo)`
`implies log(g(x)) in R`
Hence, `f(x)` is an onto function.
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